Let $f:X\to X$ be a continuous map on compact metric space $(X, d)$ with $diam(X)= 1$. A sequence $\{x_n\}_{n\in\mathbb{Z}}$ is called a $\delta$-pseudo orbit if $d(f(x_n), x_{n+1})<\delta$, for all $n\in\mathbb{Z}$. Assume that $X_f=\{ \{x_n\}: f(x_{n+1})= x_n \}$. Consider metric \begin{equation} \tilde{d}(\{x_n\}, \{y_n\})= \sum_{n\in\mathbb{Z}}\frac{d(x_n, y_n)}{2^{|n|}} \end{equation} It is easy to see that if $\tilde{d}(\{x_n\}, y_n\})<\frac{1}{2^{N+1}}$, then $x_n=y_n$ for all $|n|<N$.
In the following, I will try to show that for every $\epsilon>0$ there is $\delta>0$ such that if $(\{x_n^k\}_{n\in\mathbb{Z}})_{k\in\mathbb{Z}}\subseteq X_f$ is $\delta$-pseudo orbit for $\sigma_f:X_f\to X_f$ , then there is $z\in X_f$ with \begin{equation} \tilde{d}(\sigma_f^k(z), \{x_n^k \}_{n\in\mathbb{Z}})<\epsilon, \forall k\in\mathbb{Z}. \end{equation}
where $\sigma_f:X_f\to X_f$ is defined by $\sigma_f(\{x_n\})= \{f(x_n)\}$.
Choose $N\in\mathbb{N}$ with $\delta=\frac{1}{2^{N+1}}<\frac{\epsilon}{8}$ and assume that $(\{x_n^k\}_{n\in\mathbb{Z}})_{k\in\mathbb{Z}}\subseteq X_f$ is $\delta$-pseudo orbit for $\sigma_f:X_f\to X_f$.This means that for all $k\in\mathbb{Z}$, \begin{equation} \sum_{n\in\mathbb{Z}}\frac{d(f(x_n^k), x_n^{k+1})}{2^{|n|}}<\frac{1}{2^{N+1}} \end{equation} Hence $f(x_n^k)=x_n^{k+1}$ for every $n$ with $|n|<N$ and all $k\in\mathbb{Z}$. This implies that $x_n^{k}= f^k(x_n^0)$ if $|n|<N$. Take $z= \{x_n^0\}_{n\in\mathbb{Z}}\in X_f$. Then \begin{equation} \tilde{d}(\sigma_f^k(\{x_n^0\}), \{x_n^k\})= \tilde{d}(\{f^k(x_n^0)\}, \{x_n^k\})= \end{equation} \begin{equation} \sum_{n=N}^\infty \frac{d(f^k(x_n^0), x_n^k)}{2^{|n|}}+ \sum_{n=\infty}^{-N} \frac{d(f^k(x_n^0), x_n^k)}{2^{|n|}}\leq \sum_{|n|>N}\frac{diam(X)}{2^{|n|}}<\epsilon. \end{equation} Please help me to know that my proof is true?